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Maximum and Minimum Values of a Function

CSCA Maximum and Minimum Values of a Function study guide organized around the publicly available CSCA syllabus. Practice Mathematics questions on aicsca.com.

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Maximum and Minimum Values of a Function

1. Concept

The **Maximum and Minimum Values** (collectively called **Global Extrema**) describe the boundaries of the function's output range over its entire domain or a specific interval.

Let the domain of $y=f(x)$ be $D$:

* **Maximum Value**: If there exists $x_0 \in D$ such that $f(x) \le f(x_0)$ for **all** $x \in D$, then $y_{max} = f(x_0)$.

* **Minimum Value**: If there exists $x_0 \in D$ such that $f(x) \ge f(x_0)$ for **all** $x \in D$, then $y_{min} = f(x_0)$.

**Key Distinction**:

* **Global Extrema**: The highest/lowest points over the *entire* interval. Unique.

* **Local Extrema**: The highest/lowest points in a *local neighborhood* (peaks and valleys). Can be multiple.

* Global extrema occur at either local extrema or endpoints.

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2. Common Methods

#### (1) Graphical Method

Sketch the graph and observe the highest and lowest points.

#### (2) Completing the Square (For Quadratics)

For $f(x)=a(x-h)^2+k$ on a closed interval $[m, n]$:

1. Check if the vertex $x=h$ lies inside $[m, n]$.

2. If yes, compare $f(h)$, $f(m)$, and $f(n)$.

3. If no, only compare endpoints $f(m)$ and $f(n)$.

#### (3) Derivative Method (General Approach)

To find extrema of a differentiable function on $[a, b]$ (The **Closed Interval Method**):

1. **Differentiate**: Find $f'(x)$.

2. **Find Critical Points**: Solve $f'(x)=0$ inside the interval.

3. **Evaluate**: Calculate $f(x)$ at all **critical points** AND **endpoints** $a, b$.

4. **Compare**: The largest value is the Maximum; the smallest is the Minimum.

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3. Classic Examples

**Example 1: Quadratic on a Closed Interval**

Find max and min of $f(x)=x^2-4x+5$ on $[0, 3]$.

**Solution**:

1. **Vertex**: $f(x) = (x-2)^2 + 1$. Axis $x=2$.

2. **Check**: $x=2$ is inside $[0, 3]$.

3. **Evaluate Points**:

* Left Endpoint: $f(0) = 5$

* Vertex: $f(2) = 1$

* Right Endpoint: $f(3) = 2$

4. **Result**: Max is $5$, Min is $1$.

**Example 2: Derivative Method**

Find max and min of $f(x)=x^3-3x+1$ on $[-2, 2]$.

**Solution**:

1. $f'(x)=3x^2-3$.

2. Set $f'(x)=0 \Rightarrow x=\pm 1$. Both in $[-2, 2]$.

3. **Evaluate**:

* $f(-2) = -1$

* $f(-1) = 3$

* $f(1) = -1$

* $f(2) = 3$

4. **Result**: Max is $3$, Min is $-1$.

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4. Common Pitfalls

* **Forgetting Endpoints**: In the derivative method, students often find local peaks/valleys but forget to check if the endpoints are higher or lower.

* **Ignoring Interval**: For quadratics, blindly using the vertex formula without checking if the vertex is actually within the given domain.

* **Confusing Local vs. Global**: A local maximum is not necessarily the global maximum.